The value of the limit of frac{x^3 - x^2 - 18 | vedclass

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Question

(D) Let $y = \lim_{x 	o 3} \frac{x^3 - x^2 - 18}{x - 3}$.Substituting $x = 3$,we get $\frac{3^3 - 3^2 - 18}{3 - 3} = \frac{27 - 9 - 18}{0} = \frac{0}

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(D) Let $y = \lim_{x 	o 3} \frac{x^3 - x^2 - 18}{x - 3}$.Substituting $x = 3$,we get $\frac{3^3 - 3^2 - 18}{3 - 3} = \frac{27 - 9 - 18}{0} = \frac{0}{0}$ form.Applying $L'	ext{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:$y = \lim_{x 	o 3} \frac{\frac{d}{dx}(x^3 - x^2 - 18)}{\frac{d}{dx}(x - 3)}$$y = \lim_{x 	o 3} \frac{3x^2 - 2x}{1}$Substituting $x = 3$:$y = 3(3)^2 - 2(3) = 3(9) - 6 = 27 - 6 = 21$.

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is

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